March 23, 2015

Today:Warm-Up

Recognizing & Factoring Perfect Square Trinomials Factoring (ax2 + bx + c) Trinomials

Review of all other factoring methods/test review Reminder: Khan Academy Due Wednesday

Class Work

Complete Factoring Guide Available

@v6Math

Notes:First Factoring Test This Friday

~ Not all factoring methods will be on

first test. The review document lists the

methods on this test.

Warm-Up: Factoring Practice(8)

3. 4x³ - 4x

1. (x+9)(x-8) 2. 36x² - 25 2. (6x - 5)(6x + 5)

3. 4x(x+1)(x-1) 4. 21x³y + 28x²y²

7. x3 + 2x2 + 3x + 6

1. x² + x - 72

4. 7x²y (3x+4y)

Factor each expression completely:

5. (x + 4)² 5. (x² + 8x + 16)

The polynomial in # 2 is called a ______ polynomial

The polynomial in # 5 is called a ______ polynomial

6. (3a – 2b)²

The polynomial in # 6 is called a ______ polynomial

6. (9a² – 12ab - 4b²)

Class Notes Section of your Notebook:

8. x² - 9x + 12 Prime7. (x + 2)+ (x2 + 3)

Factoring Perfect Square TrinomialsLet's look at #'s 5 & 6 from the warm – up: 5. (x + 4)²

When multiplied, we find that (x + 4) (x + 4)

are factors of (x² + 8x + 16). This type of polynomial is known as a Perfect Square Trinomial

1. PST's have a square in the first & third terms: The first & third terms are always positive.

2. The factors of PST's are always either the square of a sum (x + y)², or the square of a difference (x - y)²

3. We arrive at the trinomial by performing a

" square, double, square" on the factor. To factor

then, we do the opposite, which is a sq. root, halve, sq. root. to arrive at the factors.

Recognizing & Factoring Perfect Square Trinomials

Expand (Multiply):(x + 2)2We'll start at the end and find the trinomial first :

We use square, double, square to arrive at: x2 + 4x + 4

Our task today, then, is to determine whether this trinomial can be factored into either the square of a sum or the square of a difference.

sq. root(x) + half(2) + sq. root(2)x2 + 4x + 4Working the other way, the opposite of square, double, square is:

The middle term is twice the product of

the square root of the first term and the

square root of the third term.

Method # 1: A trinomial is a PST, if:Check: Is this a PST? x2 + 4x + 4Method #2: A trinomial is a PST, if: b 2

2= A*C

Using Method 2: x2 + 4x + 4 = 4 2 4 = 1*4

2PST Confirmed

(3x - 4)2

Is this a perfect square trinomial?sq. root (3x) + half (12) + sq. root(4)

Always be aware of possible perfect square trinomials

(9x2 -24x + 16)

What's the Point? If you can identify a perfect square trinomial, it makes factoring very easy!

Recognizing & Factoring Perfect Square Trinomials

Yes, b = acAnd the factored form is?= A*C

b 2

2Use method 2:

Factor the following: (x² - 16x + 64) = (x - 8)²(x² + 8x + 16) = (x + 4)²(x² - 15x + 36) Is not a sp. product. Can it be factored?

Last trinomial... 9y3 + 12x2 +

4xIs it a PST? Factored Form?

When Factoring:

1. Look for the GCF

2. Look for special cases.

a. difference of two squares

b. perfect square trinomial

3. If a trinomial is not a perfect square, look for two different binomial factors.

8t4 – 32t3 + 40t

= 8t(t3 - 4t + 5)

4x2 – 9y2

=(2x)2 – (3y)2

x2 + 8x + 16

= x2 + 8x + 42 = (x + 4)2

x2 + 11x – 10

= (x + 10)(x – 1)

Factoring 2nd degree trinomials

with a leading coefficient > 1

Factoring (ax2 + bx + c) Trinomials

Factoring Trinomials

Use any of the following methods to take the

guesswork out of factoring trinomials.

It would be a good idea to write the steps down

once, as they are easy to forget when away

from class

You can use these steps for any ax2 + bx + c

polynomial, and for any polynomial you are

having difficulty factoring.

Method #1: Grouping

Step

1Multiply the leading coefficient and the constant

term

Method #1: Grouping

Step 2Find the two factors of 24 that add to

the coefficient of the middle term.

Notice the 'plus, plus' signs in the

original trinomial.

Factors of 24:

1 24

2 12

3 8

4 6

Our two factors are 4 &

6

Step 3Re-write the original trinomial

and replace 10x with 6x + 4x.

3x2 + 6x + 4x + 8Step

4 Factor by Grouping

Step

5 Factor out the GCF of each pair of

terms

After doing so, you will have...

Step 6 Factor out the common binomial, check that no further factoring is

possible, and the complete factorization is..

Practice: Factor by GroupingFactor: 6x3 + 14x2 +

8x In this case, step 1 is...

and we are left with.. 2x(3x2 + 7x + 4)

1 12

2 6

3 4

Multiply the leading coefficient and the constant

termFactors of 12: Our two factors are 3 &

4Re-write the original trinomial

and replace 7x with 3x + 4x.

2x(3x2 +3x + 4x +

4)Factor by Grouping2x(3x2 +3x) + (4x + 4)

=

2x • 3x(x +1) + 4(x + 1)

=2x (x +1)(3x +

4)

Method #2: The Box Method

Using the Box Method to factor (ax2 + bx + c)

Trinomials

9x3 + 12x2 + 4x

As usual, we

are looking

for factors

that add to

'b', and

multiply to

'ac'

Is there a GCF

to Factor?x(9x2 + 12x + 4)

3x,3x

9x, x

4,1:

2,2

3x 3x

1.Draw binomials with correct signs

4

1x( + )( +

)2. Multiply diagonally to mentally check, or fill in the binomial.

3x 4 3x 13x 2 3x 2

x( 3x+2)(3x +2)or,The correct factorization is:

x( 3x + 2)2

Factors

of 'a'

2

2

Practice: Factor using the Box Method10x2 + 21x -

10

10,-1: -

10,1

5,-2: -5,2

10x,x

5x,2x

Factors

of 'a'

Draw binomials with correct

signs( + )( +

)

( 2x+5)(5x- 2)

Method #3: Trial & Error

Factor the polynomial 25x2 + 20x + 4.

Possible factors of 25x2 are {x, 25x} or {5x, 5x}.

Possible factors of 4 are {1, 4} or {2, 2}.

We need to try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors.

Keep in mind that, because some of our pairs are not identical factors, we may have to switch some pairs of factors and make 2 attempts before we can definitely decide a particular pair of factors will not work.

Method #3: Trial & Error

We are looking for a combination that gives the sums to the middle term and are factors of the last term

{x, 25x} {1, 4} (x + 1)(25x + 4) 4x 25x 29x

(x + 4)(25x + 1) x 100x 101x

{x, 25x} {2, 2} (x + 2)(25x + 2) 2x 50x 52x

{5x, 5x} {2, 2} (5x + 2)(5x + 2) 10x 10x 20x

Method #3: Trial & Error

Class Work: 3.10